| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Joint probability of separate processes |
| Difficulty | Moderate -0.3 This is a straightforward application of Poisson approximation to binomial with standard parameters (n=120, p=0.075 gives λ=9). Part (a) requires calculating P(X>3) using tables, and part (b) simply squares this probability. The question is slightly easier than average because it explicitly tells students to use an approximation, requires only table lookup (not derivation), and involves basic probability multiplication for independent events. |
| Spec | 2.04d Normal approximation to binomial5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim B(120, 0.075)\) | B1 | Writing or use of \(B(120, 0.075)\); may be implied by using \(Po(9)\) or \(N(9, 8.325)\) |
| Approximated by \(Po(9)\) | M1A1 | 1st M1 writing or use of Poisson; 1st A1 writing or use of \(Po(9)\) |
| \(P(X > 3) = 1 - P(X \leq 3)\) | M1 | 2nd M1 for writing or using \(1 - P(X \leq 3)\); may be implied by awrt 0.972 using normal approximation |
| \(= 1 - 0.0212 = 0.9788\) | A1 | awrt 0.979 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{at least 4 defective in each box}) = P(X>3) \times P(X>3)\) | M1 | \((\text{their (a)})^2\) or \(0.979^2\) or \(0.9788^2\) or \(0.98^2\) |
| \(= 0.9788^2 = 0.95804944\) | A1 | awrt 0.958 |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(120, 0.075)$ | B1 | Writing or use of $B(120, 0.075)$; may be implied by using $Po(9)$ or $N(9, 8.325)$ |
| Approximated by $Po(9)$ | M1A1 | 1st M1 writing or use of Poisson; 1st A1 writing or use of $Po(9)$ |
| $P(X > 3) = 1 - P(X \leq 3)$ | M1 | 2nd M1 for writing or using $1 - P(X \leq 3)$; may be implied by awrt 0.972 using normal approximation |
| $= 1 - 0.0212 = 0.9788$ | A1 | awrt 0.979 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{at least 4 defective in each box}) = P(X>3) \times P(X>3)$ | M1 | $(\text{their (a)})^2$ or $0.979^2$ or $0.9788^2$ or $0.98^2$ |
| $= 0.9788^2 = 0.95804944$ | A1 | awrt 0.958 |
---\begin{enumerate}
\item The probability of an electrical component being defective is 0.075 The component is supplied in boxes of 120\\
(a) Using a suitable approximation, estimate the probability that there are more than 3 defective components in a box.
\end{enumerate}
A retailer buys 2 boxes of components.\\
(b) Estimate the probability that there are at least 4 defective components in each box.\\
\hfill \mbox{\textit{Edexcel S2 2012 Q5 [7]}}